# Horizontal Projectile Motion Speed

1. Apr 23, 2010

1. The problem statement, all variables and given/known data

A projectile is fired horzontally from a height of 1.5 m with a velocity of 100 m/s [E]. Ignoring any effects of friction....

a) How much time will elapse between the projectile being fired and hitting the ground?

b) What will be the horizontal distance from the point of firing to the point at which the prokectile lands?

c) What will be the speed of the projectile when it lands?

d) What will be the angle at which the projectile strikes the ground?

2. Relevant equations

3. The attempt at a solution
a) I'm thinking about using the equation v2 = v1 +-g(delta)t

b) (delta)d = vt is what I was thinking.

c) I attempted this one so many times, it's worth the most marks of them all, so it makes me think I need to use multiple equations.

d) not a clue

This is a correspondance physics course and the review I recieved had this projectile motion question on it, which is normally a good sign that there will be a similar question on the exam. The thing is, this grade 11 physics course did not cover horizontal projectile motion at all, and google is leading me astray with forumlas I've never even seen. I've been giving a list of formulas at the beginning of the practice exam, and I just can't determine at all which is the proper one to use. I really wanna ace my physics exam, but this is the only question I can't even begin to comprehend. The kinematics/motion on a plane unit was very scarce about these problems. Any help would be appreciated. Thanks.

2. Apr 24, 2010

### ehild

It is very important to know that force, acceleration, velocity and displacement are all vectors and the components of these vectors can be treated separately. In case of a horizontal projector, the only force is gravity and its points vertically downward. The acceleration has the same direction as the force: downward, and its magnitude is g. The projectile has got an initial horizontal velocity vo: as there is no horizontal force, the projectile keeps this horizontal velocity during all its motion. You can treat the projectile as if it did two separate motions, one along the vertical, the other in horizontal direction, and the resultant of the horizontal and vertical motion is how the projectile actually moves.

If x is the horizontal axis of the coordinate system, and the y axis points vertically downward, the initial position of the projectile is x(0)=0, y(0)=0,
the initial velocity components are vx0=150 m/s,
vy0=0,
and the acceleration is 0 in the x direction and g in the y direction.

Write the velocity components and the coordinates in terms of time for both the horizontal and vertical motion:

vx(t)=vx0
vy(t)=gt

x=vx0t
y=g/2 t^2.

At the ground, where the projectike lands, the y coordinate is yfinal=h=1.5 m. You can calculate the time of landing tL from that.
During the time of falli, the projectile moves with constant horizontal velocity so it lands at the horizontal distance d=xfinal=vx0tL.

You can find the vertical velocity component at the landing time. The total velocity at landing is the vector sum of the horizontal and vertical velocity components, and the speed is the magnitude of the velocity vector. v(final)=sqrt(vx^2+vy^2). The tangent of the angle of the velocity vector at which the projectile hits the ground is vy/vx.

ehild

Last edited: Jun 29, 2010
3. Apr 24, 2010

a) (delta)t = (delta)d/(delta)v..............(delta)t=0.015(s)

b) (9.8/2)(0.015) = 0.0735m. Therefore the projectile will have gone 0.0735m before hitting the ground

c) v=d/t; 0.0735/0.015 = 4.9. When the projectile lands it will have a speed of 4.9m/s

d) tan=opp/adj = 1.5/0.0735 = 20.408
Therefore the angle at which the projectile strikes the ground is 20.4 degrees.

How's that? Am I closing in on a possible solution? Thanks again for the help.

Last edited: Apr 24, 2010
4. Apr 24, 2010

or should c be square root(4.9^2)(9.8^2) = squareroot(24)(96) = 48 m/s?

Please keep in mind I've never solved a problem like this one yet. All help is highly appreciated :)

Last edited: Apr 24, 2010
5. Apr 24, 2010

### ehild

No, it is wrong, but I can not explain if I do not know what you have learnt so far.

ehild

6. Apr 24, 2010

Acceleration was taught thoroughly, vectors has a couple lessons. Learned about 1D and 2D vectors, vector addition and vector subtraction. There wasn't a single question structured like this. I'm giving a list of equations to use. They are as follows:

Gravity 9.8m/s^2
Snell's Law
Thin Lense Equation
P=IV
V=IR
F=MA
v=332+(T-0(degrees)C))(0.59)

These are the equations listed to use during the exam review.

7. Apr 24, 2010

The exam review is structured to be out of 100, this question has a total of 16 marks to it, so the real deal with a similar question will be worth the same I guess. It's imperative I understand this, it's just so hard because I have no idea what to do. Again, any and all help is very appreciated.

8. Apr 25, 2010

### ehild

Well, so you have learnt about gravity. Try to answer: A body falls from 1.5 m height. How long does it take to reach the ground?

ehild

9. Apr 25, 2010

### ehild

Why do you think so? I asked time. It is measured in seconds.

Look after the definition of acceleration, and the equations for "motion with uniform acceleration"

ehild

10. Apr 25, 2010

sqrt(2)(1.5)/9.81 = 0.5530 seconds

11. Apr 25, 2010

Yea I posted that first one without thinking at all haha. Sorry I've had my head in the books for days. I'm a little on edge.

12. Apr 25, 2010

### ehild

Yes but write it out as t=sqrt((2)(1.5)/9.81).

The motion of the projectile (say, a cannon ball) can be decomposed into a vertical motion and a horizontal one. The vertical motios is like a free fall, with acceleration g. The horizontal motion is with constant velocity of 100 m/s.

You just calculated that 0.553 s is needed to reach the ground from 1.5 m height. During that time, the ball is in air, and flying also horizontally. How far does it reach in horizontal direction in 0.553 s, if the horizontal velocity is 100 m/s ?

ehild

13. Apr 25, 2010

I believe this is where I use the equation: d = vt + 1/2at^2

d = (100)(0.55) + 1/2(100/0.55)(0.55)^2

d = 55 + 27.5

d = 82.5 m

Last edited: Apr 25, 2010
14. Apr 25, 2010

Just wanna say; thanks so much for your time/patience/help with all of this ehild. It means alot to me :)

15. Apr 25, 2010

The only way I remember how to calculate speed is v =d/t
If the above answer is correct then the speed should be:

v = 82.5/0.55
v = 150 m/s

Althugh this seems almost too easy considering how part c is worth the most marks. I found another equation that I think would be appropriate for this question.

v = vi^2 + (2)(9.8)
vi = initial velocity which should be 0 if its fired from a projectile meaning its at rest.

v= 0^2 + 19.6
take the sqrt of 19.6
v = 4.43 m/s

Am I on the right track now? Thanks again.

Last edited: Apr 25, 2010
16. Apr 25, 2010

tan-1(1.5/0.55 = 69.86 = 70 degrees

So the angle at which it lands is 70 degrees. I hope I'm starting to get the idea :)

17. Apr 25, 2010

### ehild

No. The ball or anything moves with constant velocity in the horizontal direction. The horizontal acceleration (a) is zero. From where did you get that 100/0.55 for the acceleration? The ball started with 100 m/s horizontal velocity, and kept it, as there is no force acting horizontally. How far does a ball go in 0.553 s with a constant velocity of 100 m/s?

Just think: you drive a car with 100 km/h, how do you calculate the distance you travel in 0.5 hours?

ehild

18. Apr 25, 2010

A car would travel 50km's going 100km/h in 0.5 hours. That pretty basic. So the answer for distance travelled is simply (100m/s) (0.55s) = 55 meters?

19. Apr 26, 2010

### ehild

YES! You got that the projectile reaches the ground 55 m away. Go ahead. You have to calculate the velocity components at the impact. The horizontal velocity is the same during the whole flight, 100 m/s.
The vertical component of velocity linearly increases with time, and it was zero at start, so the vertical velocity is gt at the impact.

If you know the horizontal and vertical velocity components, you get the magnitude of velocity (speed) with Pythagoras' law:
v=sqrt(vx2+vy2).

The tangent of the angle the velocity vector encloses with the ground is tan(alpha) = vy/vx.

ehild

20. Apr 26, 2010